\(\int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx\) [824]

Optimal result

Integrand size = 23, antiderivative size = 22 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=(-a-b x)^{-n} (a+b x)^n \log (x) \]

[Out]

(b*x+a)^n*ln(x)/((-b*x-a)^n)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {23, 29} \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=\log (x) (-a-b x)^{-n} (a+b x)^n \]

[In]

Int[(a + b*x)^n/(x*(-a - b*x)^n),x]

[Out]

((a + b*x)^n*Log[x])/(-a - b*x)^n

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps \begin{align*} \text {integral}& = \left ((-a-b x)^{-n} (a+b x)^n\right ) \int \frac {1}{x} \, dx \\ & = (-a-b x)^{-n} (a+b x)^n \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=(-a-b x)^{-n} (a+b x)^n \log (x) \]

[In]

Integrate[(a + b*x)^n/(x*(-a - b*x)^n),x]

[Out]

((a + b*x)^n*Log[x])/(-a - b*x)^n

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.79 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64

method	result	size
risch	\(\ln \left (x \right ) {\mathrm e}^{-i n \pi \left (\operatorname {csgn}\left (i \left (b x +a \right )\right )^{3}-\operatorname {csgn}\left (i \left (b x +a \right )\right )^{2}+1\right )}\)	\(36\)

[In]

int((b*x+a)^n/x/((-b*x-a)^n),x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(-I*n*Pi*(csgn(I*(b*x+a))^3-csgn(I*(b*x+a))^2+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.36 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=e^{\left (i \, \pi n\right )} \log \left (x\right ) \]

[In]

integrate((b*x+a)^n/x/((-b*x-a)^n),x, algorithm="fricas")

[Out]

e^(I*pi*n)*log(x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.72 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=\begin {cases} e^{- i \pi n} \log {\left (-1 + \frac {b \left (\frac {a}{b} + x\right )}{a} \right )} & \text {for}\: \left |{\frac {b \left (\frac {a}{b} + x\right )}{a}}\right | > 1 \\e^{- i \pi n} \log {\left (1 - \frac {b \left (\frac {a}{b} + x\right )}{a} \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**n/x/((-b*x-a)**n),x)

[Out]

Piecewise((exp(-I*pi*n)*log(-1 + b*(a/b + x)/a), Abs(b*(a/b + x)/a) > 1), (exp(-I*pi*n)*log(1 - b*(a/b + x)/a)
, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.27 \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=\left (-1\right )^{n} \log \left (x\right ) \]

[In]

integrate((b*x+a)^n/x/((-b*x-a)^n),x, algorithm="maxima")

[Out]

(-1)^n*log(x)

Giac [F]

\[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=\int { \frac {{\left (b x + a\right )}^{n}}{{\left (-b x - a\right )}^{n} x} \,d x } \]

[In]

integrate((b*x+a)^n/x/((-b*x-a)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)^n/((-b*x - a)^n*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(-a-b x)^{-n} (a+b x)^n}{x} \, dx=\int \frac {{\left (a+b\,x\right )}^n}{x\,{\left (-a-b\,x\right )}^n} \,d x \]

[In]

int((a + b*x)^n/(x*(- a - b*x)^n),x)

[Out]

int((a + b*x)^n/(x*(- a - b*x)^n), x)